Generalize the Monty Hall problem where there are $n \geq 3$ doors, of
which Monty opens $m$ goat doors, with $1 \leq m \leq n$.
Original Monty Hall Problem: There are $3$ doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don’t want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is Door $1$. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening Door $2$and Door $3$, he chooses Door $2$ and Door $3$ with equal probability .
Find the probability that the strategy of always switching succeeds, given that Monty opens Door $2$.
My approach:
Let $C_i$ be the event that car is behind the door $i$,
$O_i$ be the event that Monty opened door $i$ and $X_i$ be the event that intially I chose door $i$. Here $i=1,2,3,…,n$.
Let's start with case where I chose $X_1$. Then:
$P(O_{j_1, j_2, …, j_m}|C_1, X_1) = {{n-1}\choose{m}}(\frac{1}{n-1})^m$, here $j \in$ {$m$ doors out of $n-1$, i.e., exclude Door$1$ }
$P(O_{k_1, k_2, …, k_m}|C_t, X_1) = {{n-2}\choose{m}}(\frac{1}{n-2})^m$,
here $k \in$ {$m$ doors out of $n-2$, i.e., exclude Door$1$ & Door$t$}, $t \in$ {$2,3, …, n$}
Also, $P(C_r|X_s) = \frac{1}{n}$, here $r,s \in$ {$1,2,…,n$}
Probability of winning by switching is,
$$P(C_3 | O_{k_1, k_2, …, k_m}, X_1) = \frac{P(O_{k_1, k_2, …, k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{m-doors}|X_1)}$$
$$= \frac{P(O_{k_1, k_2, …, k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{j_1, j_2, …, j_m}|C_1, X_1).P(C_1|X_1) + \sum_{t=2}^n(P(O_{k_1, k_2, …, k_m}|C_t, X_1).P(C_t|X_1))}$$
$$ = \frac{{{n-2}\choose{m}}(\frac{1}{n-2})^m.\frac{1}{n}}{(\frac{1}{n}).({{n-1}\choose{m}}(\frac{1}{n-1})^m + {{n-2}\choose{m}}(\frac{1}{n-2})^m.(n-1))}$$
$$= \frac{(n-m-1)(n-1)^{m-1}}{(n-2)^m + (n-m-1)(n-1)^m}$$
However, the correct answer is $\frac{(n-1)}{(n-m-1).n}$. Any insights to what I have done wrong.
Best Answer
If your initial door is the car, the switching strategy fails.
$\frac{n-1}{n}$ of the time, your initial door is not the car. Then excluding your initial door and the $m$ goat doors, the car must be in one of the $n-m-1$ remaining doors, and presumably the switching strategy picks uniformly at random. So, the probability that switching succeeds is $\frac{n-1}{n} \cdot \frac{1}{n-m-1}$.